Don't worry about maintaining the integrity of Node A (or any of the nodes) for now. First find the total amount of current flowing in the battery. Once you know that you know the current in some of the resistors and can walk the voltage drops from a known voltage (one side or the other of the battery) to one of the nodes to find the voltage at that node. Do that for as many nodes as you can. Then use that new knowledge to tease out the rest of the voltages.

For instance, if you already knew the voltage at B (say I told you what it was), could you use that information, without knowing any of the currents in the circuit, to figure out the voltage at Node A?

"Now you have three resistors in series, 200, 100 and 100 ohms."
Now you have two resistors in series, 200, 200 ohms.And Vb between them.
So Vb is half the supply voltage Vb= 6V.
Vc is half this 6V, so Vc=3V.
Va it is between R1=100ohms and R2=300ohms:
R1 have a quarter and R2 have three-quarters form a voltage difference between V+ and Vb, that is 6V.
6/4V on R1 => Va=12V-1.5V

Please check the solving, I wrote in haste.But this is not the way to solve. you have to apply the formulas and learned rules.

"Now you have three resistors in series, 200, 100 and 100 ohms."
Now you have two resistors in series, 200, 200 ohms.And Vb between them.
So Vb is half the supply voltage Vb= 6V.
Vc is half this 6V, so Vc=3V.
Va it is between R1=100ohms and R2=300ohms:
R1 have a quarter and R2 have three-quarters form a voltage difference between V+ and Vb, that is 6V.
6/4V on R1 => Va=12V-1.5V

Please check the solving, I wrote in haste.But this is not the way to solve. you have to apply the formulas and learned rules.